Question

Point $P$ divides the line segment joining the points $A(-1,3),B(9,8)$ such that $\frac{AP}{PB}=\frac{k}{1}$. If $P$ lies on the line $x-y+2=0$, find the value of $k$.

Answer 1

Using the section formula, if a point $(x,y)$ divides the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$, then

$(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Given: $\frac{AP}{BP}=\frac{k}{1}$

So, $P$ divides the line segment joining the points $A(-1,3)$ and $B(9,8)$ in the ratio $k:1$

Using the section formula, we get

Coordinates of $P=(\frac{9k-1}{k+1},\frac{8k+3}{k+1})$

Since $P$ lies on the line $x-y+2=0$, so

$\Rightarrow \frac{9k-1}{k+1}-\frac{8k+3}{k+1}+2=0$

$\Rightarrow \frac{9k-1-8k-3}{k+1}+2=0$

$\Rightarrow \frac{k-4}{k+1}=-2$

$\Rightarrow k-4=-2k-2$

$\Rightarrow 3k=2$

$\therefore k=\frac{2}{3}$